How do you find the limit of (sqrt(x^2+11)-6)/(x^2-25) as x approaches 5?

Jamya White

Jamya White

Answered question

2023-03-06

How to find the limit of x 2 + 11 - 6 x 2 - 25 as x approaches 5?

Answer & Explanation

nanaandhachi9ur

nanaandhachi9ur

Beginner2023-03-07Added 3 answers

Answer: 1 12
Consider how evaluating the limit now would be impossible because the fraction's denominator would be 0.
When dealing with radical expressions and other general nastiness, multiplying by a conjugate is a good strategy. We can try multiplying this by the numerator's conjugate.
= lim x 5 x 2 + 11 - 6 x 2 - 25 ( x 2 + 11 + 6 x 2 + 11 + 6 )
In the numerator, we can recognize that this is really the difference of squares pattern in reverse, where ( a + b ) ( a - b ) = a 2 - b 2 .
= lim x 5 ( x 2 + 11 ) 2 - 6 2 ( x 2 - 25 ) ( x 2 + 11 + 6 )
= lim x 5 x 2 + 11 - 36 ( x 2 - 25 ) ( x 2 + 11 + 6 )
= lim x 5 x 2 - 25 ( x 2 - 25 ) ( x 2 + 11 + 6 )
Notice the x 2 - 25 terms in the numerator and denominator will cancel.
= lim x 5 1 x 2 + 11 + 6
We can now calculate the limit by substituting 5 for x.
= 1 25 + 11 + 6 = 1 36 + 6 = 1 12
Even though the point at x = 5 is undefined, it is very close to 1 12 = 0.08 3 ¯ .

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